$\sum_{k\in\mathbb{Z}^+}\:\frac{\sin(2kx)}{k}$ serisini hesaplayın

Question

$$\sum_{k\in\mathbb{Z}^+}\:\frac{\sin(2kx)}{k}\:\:\:,\:\:\:x\in(0,\pi)$$

Serisini hesaplayın.

Answer 1

Serimiz :

$$\sum_{k\in\mathbb{Z}^+}\:\frac{\sin(2kx)}{k}$$

Sinüsü açalım.

$$\sum_{k\in\mathbb{Z}^+}\:\frac{e^{2ikx}-e^{-2ikx}}{2ik}$$

Sadeleştirelim.

$$\frac{1}{2i}\Bigg[\sum_{k\in\mathbb{Z}^+}\:\frac{e^{2ikx}}{k}-\sum_{k\in\mathbb{Z}^+}\:\frac{e^{-2ikx}}{k}\Bigg]$$

Artık serileri hesaplayabiliriz.

$$\frac{1}{2i}\bigg[-\ln(1-e^{2ix})+\ln(1-e^{-2ix})\bigg]$$

Sadeleştirelim.

$$\frac{1}{2i}\ln\bigg(\frac{1-e^{-2ix}}{1-e^{2ix}}\bigg)$$

$$\frac{1}{2i}\ln\big(-e^{-2ix}\big)$$

$$\large\color{#A00000}{\boxed{\sum_{k\in\mathbb{Z}^+}\:\frac{\sin(2kx)}{k}=\frac{\pi}{2}-x\:\:\:,\:\:\:x\in(0,\pi)}}$$